I want to show that $\dim V = \dim \phi(V)+\dim \ker \phi$.

I know this proof can be found in any linear algebra textbook. However, my question is not exactly about the proof, but on a statement I assumed to be true while attempeting to prove it myself.

In order to prove this, let $\{{e_1,\dots,e_p}\}$ be a basis of $\phi^{-1}\{0\}$ and $\{{e_1,\dots,e_p},{\bar{e_1},\dots,\bar{e_q}}\}$ be a basis of $V$. Then, $\dim \ker \phi=p$ and $\dim V = p+q$.

So it suffices to show that $\{{\phi(\bar{e_1}),\dots,\phi(\bar{e_q})}\}$ is a basis of $\phi(V)$.

I managed to show that $\{{\phi(\bar{e_1}),\dots,\phi(\bar{e_q})}\}$ spans $\phi(V)$. But I still need to prove that the set $\{{\phi(\bar{e_1}),\dots,\phi(\bar{e_q})}\}$ is linearly independent. This is how I approached the problem:

Consider $a^{i}e_i + b^{i}\bar{e_i}=0$. Assume the sum is over the whole sets, i.e, the first goes up to $p$ and the second up to $q$. Since $\{{\bar{e_1}},e_i\}$ is a basis of $V$, $a^{i}=b^{i}=0$ for all $i$.

Since $\phi(0)=0$ and $\phi$ is linear: $$ \phi(a^{i}e_i + b^{i}\bar{e_i})=\phi(0)=0 \Rightarrow a^{i}\phi(e_i) + b^{i}\phi(\bar{e_i})=0 $$

We also know that $\{{e_1,\dots,e_p},{\bar{e_1},\dots,\bar{e_q}}\}\subset \phi^{-1}(0)$, so $\phi(e_i)=0$ for all $i$. Thus, we have: $$ b^{i}\phi(\bar{e_i})=0 $$ and all the $b^{i}$'s are $0$.

However, I know I should start with this last equation and then conclude that all the $b^{i}$'s are $0$. I can reverse my reasoning and show that this also works in the other way round.

To do so, I would have to assume that $\phi(a^{i}e_i + b^{i}\bar{e_i})=\phi(0)$ implies $a^{i}e_i + b^{i}\bar{e_i}=0$. But I don't see how the second statement follows from the first.


1 Answer 1


I'm going to use Einstein summation notation since it appears that that is what you are doing.

Suppose $$b^i\phi(\overline{e}_i)=\phi(b^i\overline{e}_i)=0$$ This means that there exist $a^i$ such that $$b^i\overline{e}_i=a^ie_i$$ based on the fact that the $e_i$ span the kernel of $\phi$, and $b^i\overline{e}_i$ is in that kernel. It follows that $$b^i\overline{e}_i-a^ie_i=0$$ Hence all coefficients are $0$. In particular, $b^i=0$ for all $i$.

  • $\begingroup$ That was really clear! Thanks! $\endgroup$
    – Thiago
    Dec 21, 2014 at 22:55
  • $\begingroup$ @Thiago You're welcome. $\endgroup$ Dec 21, 2014 at 22:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .